`**) 1 + 2 + ... + n = ( n. (n + 1) )/2`
Giải:
Đặt `A = -1/3(1 + 2 + 3) - 1/4(1 + 2 + 3 + 4) - ... - 1/50(1 + 2 + ... + 50)`
`= - [ \frac{1}{3}(1 + 2 + 3) + \frac{1}{4}(1 + 2 + 3 + 4) + ... + \frac{1}{50}(1 + 2 + ... + 50) ]`
Ta có: `1 + 2 + 3 = (3.4)/2`
`1 + 2 + 3 + 4 = (4.5)/2`
`...`
`1 + 2 + ... + 50 = (50.51)/2`
`=> A = - [1/3. (3. 4)/2 + 1/4. (4. 5)/2 + ... + 1/50. (50. 51)/2]`
`= - (4/2 + 5/2 + ... + 51/2)`
`= -1/2. (4 + 5 + ... + 51)`
Đặt `B = 4 + 5 + ... + 51`
`= (51 + 4). [ (51 - 4) : 1 + 1] : 2`
`= 55. 48 : 2`
`= 55. 24`
`= 1320`
`⇒ A = -1/2. 1320 = -660`
`Vậy A = -660`
